Red Black Red Blzck Stack

Phase One

Shuffle the deck without disturbing the twenty-six-card stack on top. Someone cuts off less than half the deck, and you glimpse the next card. With that, you have the information necessary to state how many cards of each color are in the spectator's packet. Just mentally go thmugh the colors of the stack. This becomes easier if you are aware that Cards 1 through 10 of the stack include five red and five black cards, and Cards 1 through 18 contain nine of each color. See Note 1 for further helpful statistics

Phase Two

Add five red cards and five black cards from the unstacked portion to the top of the stack. These added cards may be in any order. The spectator now cuts a larger packet—as many as thirty-six cards—and yet you aa* able to tell the audience (after glimpsing the top card of the talon) how many red or black cards are in the packet. Just add five to those in that portion of the stack Phase T>*ree

This time, the well-shuffled unstacked half is moved to the top. A spev-tator cuts off more than twenty-six cards and you again divine how many of each color he holds. Since the unstacked half consists of fourteen black cards and twelve red cards, all you need to do is make your ^btums as if the stack were on top, and then add fourteen to the number of black cards and twelve to the number of reds.

Cut the deck, separating the stacked half from the uns^ked cardv

Have a spectator shuffle the unstacked half while you ^^^ half a fa 1st- shuffle. Take the portion shuffled by the spectato and, pretend ing to weigh , in vour hand say, "Thereonfourteenth***« Un*r

250 / JUAN TAMARIZ

» iho r uds to prove the truth of your state-

the pto of nmning «*«#>»« « ^ Green's or Moizinser'»,

£ cards. Set the black cards on top.

^Perfect faro shuffle, interlace the two twenty-six-card packets/ ,eavig from the top and the K* on the bottom. Follow this with a couple of false cuts. The spectator now cuts anywhere he withe* and keep* the cut-off portion. Glimpse the top two cards of the talon, to make sure one of them belongs to the stacked half. If the card glimpsed from the stacked half is from those between the 44 (1) and the 8v (14), you'll know that the spectator's packet consists of as many red cards as it would if you were performing Phase One, where you were handling (inly the stacked portion- On the other hand, if the card glimpsed has a mnemonic number higher than that of the 8*, there v/ill be as many black cards as in the f tadced portion, plus fourteen. Try it out, cards in hand; Thus, once you've glimpsed the top two cards of the talon, you only remark on the number of red cards or the number of black cards, as convenient.

Variant I

To eliminate the faro shuffle and strengthen the effect in Phase f ive, you could do this. Once you have secretly separated the red and black cards at the end of Phase Four, set the packet of twelve urotacked red cards besjde the stacked portion on the table. Both should be face down. Have someone nffle shuffle these packets together while you shuffle, overhand fashion, the remaining fourteen black cards. Take the shuffled black pack/ft into your nght hand and the spectator's shuffled one into your left

Run three black cards from the right hand's packet onto the left hand's and throw the re* of the right hand's cards under the left's. You now ' : > rrorr •• top dov.rt ^

three blzck card*, twelve red cards rnfcrnrungied with the twenty-m card stack, and eleven black cards fF>g 56).

V uffe the deck without <S*u*ing the top forty««

J» too fartyone cards, avid

by *« tp«^ ^ t 1 veoiorkf without fere* and with *

by *« tp«^ ^ t 1 veoiorkf without fere* and with *

you know the bottom card of the- stacked half in tiv.t how m;my black card* th*< p*ket contains, Mak* y Jr7 l 'iT' ^ ^ had only the stac kx'd <ards, and then add tim* mo* bJaA ,7' ^ '' In other words, you figure out how many bU k «„j* a* ,n a* if you were performing Pha<* ()ne, but then add tf* « n, f, " ' 1

Being able to U ll how many card» of , ^ nAo, „„. packet cut by someone, nit» ¡ha deck ha* been shuffle ., baffles even even the most discerning.

Variant II

This version is even more incredible in its effect Two packet» are laid on the table.

One of ti.'-m consists of the stacked portion/ with five blac k cards from th*' umtacked por-tion added on top. The other packet contains the twelve unstacked red cards on top of the nine remaining bb< k < ards, The spectator riffle shuffles the packets together. lx*f s assume the card* up i/« the order thown in Bg 57, where the cards an- left unsquared to clarify fix* arrant-merit

Spread the deck face up on the table, undrr rh* preterw oi showing that the colors are well shu/fled, Look for the point whet* tr * umiKi*d red „nd black card* meet You wilJ immediately rv/fc- that th*- card ^ the red batch (the 7* iri Hg, 57; is under the (all opy r* H r first '¿rd of th#- black bat/h (the 64) is over the ^ TV-reform 'h" deck is turned face down, if th<7 cut above the 9T the r. jmUr of bi* k cards will be five more than it would have b«-"N /f the cut-oil p*'« < m^Jif up only of card* belonging to th<< half *ta< k If < w f> * the number of &d cards will be twelve more tilS0 U would have bt*n if the packet consisted only of rards from the Mf *acV fen*""-* yc* v/ill need to glimpse two or more bottom cards before you ftf.d ^ belongs to the half k, whirh is your key for detaining tf r nu/r or black or red cards in that packet4 , v

how many red and black caids are in the packet Voaof of one color or the other there are The §pect*U>n, however, r*v ^ ^

' // f^' f), y; u, fly u/dif i> - ^ ^ ^^ f! . .k> -

t/> Ml nurrbtf <* MmV <*r6%. m wrfJ m ^ ^ " ^ d lf Md frv* biari c^ lwHv* red* ^^w ^ly f^rdv from fhr ^ ^uJf

/ JUAN TAMARIZ

, cince vou, after weighing the cut-off packet in aware of this limitation bl„ck6ards." for example. Once vour hand, announce ^^^effect is over.

your Statement «proven ^ ^^ mixed wilh the extra five cards If some nni cards . F p unless tlie spectator cuts a very small convenient to ask, « off o packet fttf « H0,

-ri ^JSBUi the cards and you spread them face up andnot^inder which card of the stacked portion the last unstacked red S lie. That card is your key. Turn the deck face down and have a spectator cut off a packet. Glimpse one or more bottom cards, until you see one thitbelongs to the half stack. If the mnemonic number of that card is equal to or lower than that of the key card, it will tell you the number of black cards in the packet. If the mnemonic number is higher than your key card, it will tell you how many red cards there are. But before you get dizzy from reading all these explanations, I suggest you—oh-so-patient reader-look at the figures and practice with cards in hand, to better understand what's going on. Note I

Here are some more facts that allow you to determine quickly how many red and black cards there are in the cut-off packet when the half stack alone is on top (Phase One): If there is a black card with an even mnemonic number on the bottom of the packet, that packet has as many red cards as black cards. Thus:

Bottom card

Black

This trick can be combined with "Weighing the Cards" (p. 197) and

^ of Touch and Other Senses" (p. 204). Note ill r the^sh'm ^ bt>aUtifUl COmbination: On finishing Phase Five, with of "\y„h V ?K?Plained' you can Perform my Mnemonica variation Nuth B nd Nor Stupid", described at the beginning of Appendix I

'tfiCk'" y°U aPPr°ach lh" identification of the deck divided int^ ** °f the P^ts. You will end up with the

Portions, one consisting of the twenty-six stacked

Mnemonic a j 253

cards, and the other containing twelve red cards and fourt m separated. You may now do "Colors on Parade" (p ]8a T ,acks'colnrs could use my version of "Out of This World" title j "a o, * y"u be published in my forthcoming book Flamenco ' * t0

The Illustrated Key To The Tarot

The Illustrated Key To The Tarot

The pathology of the poet says that the undevout astronomer is mad the pathology of the very plain man says that the genius is mad and between these extremes, which stand for ten thousand analogous excesses, the sovereign reason takes the part of a moderator and does what it can. I do not think that there is a pathology of the occult dedications, but about their extravagances no one can question, and it is not less difficult than thankless to act as a moderator regarding them.

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